Operator Generic Fundamentals – Thermodynamic Cycles

Operator Generic Fundamentals 193005 – Thermodynamic Cycles
K1.01 Define thermodynamic cycle. K1.02 Define thermodynamic cycle efficiency in terms of net work produced and energy applied. K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency. K1.04 Describe the moisture effects on turbine integrity and efficiency. K1.05 State the advantages of moisture separators/repeaters and feedwater heaters for a typical steam cycle. Operator Generic Fundamentals – Thermodynamic Cycles

Terminal Learning Objectives
At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of ≥ 80 percent on the following TLOs: Explain compression processes and the laws associated with them. Apply the second law of thermodynamics to analyze real and ideal systems and components. TLO 1 is not tested by the NRC, but the concepts help better under stand pressurizer concepts as well as cycle efficiency. TLOs

Compression Processes
TLO 1 – Explain compression processes and the laws associated with them. 1.1 Describe the ideal gas laws and how to solve for an unknown pressure, temperature, or volume. 1.2 Describe the effects of pressure and temperature changes on confined fluids. TLO 1

Gas Laws ELO 1.1 – Describe the ideal gas laws and to solve for unknown variables. Because of their interrelated effect, temperature, pressure, and volume must always be specified when gases are discussed Their quantitative relationships expressed in gas laws Gas laws useful because at low pressures, all real gases behave like a perfect gas Monatomic gas behavior similar to perfect gas behavior; ideal gas law therefore accurate for predicting gas behavior Accuracy decreases with diatomic and polyatomic gases ELO 1.1

Boyle’s and Charles’s Laws
At low pressures, volume of a gas at constant pressure directly proportional to temperature of gas Boyle’s Law At low pressures, volume of a gas at constant temperature is inversely proportional to absolute pressure of gas Figure: Charles’s Law for Constant Pressure Boyle's Law: V and P Related KAs - Figure: Boyle's Law ELO 1.1

Boyle - pressure of gas expanding at constant temperature varies inversely to volume (𝑃1)(𝑉1)=(𝑃2)(𝑉2)=(𝑃3)(𝑉3)= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Charles - pressure of gas varies directly with temperature when volume is held constant, and volume varies directly with temperature when pressure held constant 𝑉 1 𝑉 2 = 𝑇 1 𝑇 2 𝑜𝑟 𝑃 1 𝑃 2 = 𝑇 1 𝑇 2 Figure: Combined Gas Law ELO 1.1

Combined gas law: 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 = 𝑃𝑉 𝑇 Temperature and Pressure MUST BE IN ABSOLUTE Stress Absolute scales for T & P Figure: PTV Diagram for Combined Gas Law ELO 1.1

Example Problem A compressor discharges into an air receiver and cycles off when the pressure in the receiver reaches 160 psia. During the compression, heat was added to the air and its temperature in the receiver is 140°F. Assuming no air loads are in service, at what temperature (°F) should the compressor restart to maintain the receiver above 150 psia? Show that this is a constant volume problem. Figure: PTV Diagram for Combined Gas Law ELO 1.1

Example Problem Solution
A compressor discharges into an air receiver and cycles off when the pressure in the receiver reaches 160 psia. During the compression, heat was added to the air and its temperature in the receiver is 140°F. Assuming no air loads are in service, at what temperature (°F) should the compressor restart to maintain the receiver above 150 psia? 𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2 The receiver volume is constant, therefore: 𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 𝑇 1 °𝑅 =460°+140℉=600°𝑅 𝑃 1 =160 𝑝𝑠𝑖 𝑃 2 =150 𝑝𝑠𝑖 160 𝑝𝑠𝑖 600°𝑅 = 150 𝑝𝑠𝑖 𝑇 2 𝑇 2 = 600× 𝑇 2 °𝑅 =562.5°𝑅 𝑇 2 °𝐹 =562.5°𝑅−460°=102.5℉ Keep in mind that as an operator, we want to make sure the compressor “kicks on” at the proper pressure, but temperature. ELO 1.1

Ideal Gas Law Combining results of Charles' and Boyle's experiments, the following is obtained using the specific volume 𝑣 = V/M : 𝑃𝑣 𝑇 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 This constant called universal gas constant and is designated by R0, thus ideal gas equation becomes: 𝑃𝑣 = 𝑅0𝑇 Pressure and temperature are absolute values ELO 1.1

Ideal Gas Law One mole of any substance is that amount having Avogadro’s Number of x 1023 atoms: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑚𝑜𝑙𝑒𝑠= 𝑀𝑎𝑠𝑠 𝑔𝑟𝑎𝑚𝑠 𝐴𝑡𝑜𝑚𝑖𝑐 𝑀𝑎𝑠𝑠 𝑔𝑟𝑎𝑚𝑠 𝑚𝑜𝑙𝑒 , or mass = (# moles)(GMW) At STP: 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑇 = 0°𝐶=460°𝑅 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃 =1 𝑎𝑡𝑚 𝑉𝑜𝑙𝑢𝑚𝑒 𝑉 = 𝑛 𝑙𝑖𝑡𝑒𝑟𝑠 𝑚𝑜𝑙𝑒 ELO 1.1

Ideal Gas Law Individual gas constant (R) obtained by dividing universal gas constant (Ro) by molecular weight (MW) of gas Units of R must always be consistent with units of pressure, temperature, and volume used in gas equation No real gases follow ideal gas law or equation completely Also keep in mind that the Ideal Gas law doesn’t really apply to our steam systems either because we operate at higher pressures than where this applies. Plus we use steam. This concept, however, can apply to things like our Hydrogen and/or Nitrogen pressure systems. ELO 1.1

Ideal Gas Law – Compression Process
ELO 1.2 – Describe the effects of pressure and temperature changes on confined fluids. Most common use of gas behavior is compression process using ideal gas approximations Process may occur at constant temperature (PV = constant), constant volume, or adiabatic (no heat transfer) Results in work performed on system and is essentially area under a P-V curve Figure: Pressure-Volume Diagram ELO 1.2

Compressibility of Fluids
Fluid is any substance that conforms to shape of its container May be either a liquid or a gas A fluid may be considered incompressible when velocity of fluid is greater than one-third speed of sound for fluid, or if liquid Treatment of a fluid considered incompressible is easy because density assumed to be constant, giving a simple relationship for state of the substance Variation of fluid density with changes in pressure a primary factor considered in deciding whether a fluid is incompressible Once substance becomes a gas, process becomes more difficult ELO 1.2

Constant Pressure Process
To determine work done in a constant pressure process, following equation used: W1-2a = P(ΔV) To determine work done in a constant temperature process, following equation used: W1-2b = T(ΔV) Figure: P-V Diagram for Gas ELO 1.2

Constant Volume Process
Work done is product of volume and change in pressure: 𝑊1−2𝑐=𝑉(Δ𝑃) Power requirement for pumps that move incompressible liquids (such as water) determined from replacing volume (V) with product of specific volume and mass: 𝑊1−2𝑐=𝑚𝑣(Δ𝑃) Taking time rate of change of both sides determines power requirements of pump Figure: P-V Diagram for Gas 𝑊 1− 2 𝑐 = 𝑚 𝑣(∆𝑃) ELO 1.2

Effects of Pressure Changes on Fluid Properties
Predominant effect of an increase in pressure in a compressible fluid such as a gas is increase in density of fluid Increase in pressure of incompressible fluid will not have a significant effect on density Increasing pressure of 100°F water from 15 psia to 15,000 psia will only increase density by approximately 6% ELO 1.2

Effects of Pressure Changes on Fluid Properties
Increase in temperature will tend to decrease density of any fluid Effect of a temperature change will depend on whether fluid is compressible If fluid is gas, it will respond as predicted by ideal gas laws A 5% increase in absolute temperature will result in a 5% increase in absolute pressure If fluid is incompressible liquid in closed container, increase in temperature could be catastrophic Rule of thumb for water in a water-solid system 100 psi increase for every 1°F increase in temperature If fluid is incompressible liquid in closed container, increase in temperature will have a tremendously greater and potentially catastrophic effect As fluid temperature increases, it tries to expand, but expansion prevented by walls of container Because fluid incompressible, results in tremendous increase in pressure for relatively minor temperature change Change in specific volume for a given change in temperature is not the same at various beginning temperatures, resultant pressure changes will vary ELO 1.2

Ideal Gas Law – Compression Process
Knowledge Check A contained fluid is heated. The resulting change in pressure will be… greater for an incompressible fluid. greater for a compressible fluid. the same for both fluids. the same for both fluids only if the volume is held constant. Correct answer is A. Correct answer is A ELO 1.2

Second Law of Thermodynamics
TLO 2 – Apply the second law of thermodynamics to analyze real and ideal systems and components. Figure: Entropy of Ice and Water TLO 2

TLO 2 2.1 Explain the Second Law of Thermodynamics using the term entropy. 2.2 Given a thermodynamic system, determine the: Maximum efficiency of the system Efficiency of the components within the system 2.3 Differentiate between the path for an ideal process and that for a real process on a T-s or h-s diagram. 2.4 Describe how individual factors affect system or component efficiency. TLO 2 - ELOs

ELO Explain the Second Law of Thermodynamics using the term entropy. It is impossible to construct a device that operates in a cycle and produces no effect other than the removal of heat from a body at one temperature and the absorption of an equal quantity of heat by a body at a higher temperature. Related KA K1.01 Explain the relationship between real and ideal processes. 1.8* 1.9* Figure: Second Law of Thermodynamics for a Heat Engine ELO 2.1

Concept of second law is best stated using Kelvin & Planck’s description: It is impossible to construct an engine that will work in a complete cycle and produce no other effect except the raising of a weight and the cooling of a heat reservoir Summary of First and Second Laws: “Not only can’t you get something from nothing, you can’t break even”. The First Law represents the energy conversion that take place place in our processes. The Second Law represents the “change in entropy”. To make the cycle more efficient, try and minimize the change in entropy in each of the four processes. Figure: Kelvin-Planck’s Second Law of Thermodynamics ELO 2.1

The Second Law of Thermodynamics is needed because the First Law of Thermodynamics does not completely define the energy conversion process The first law is used to relate and evaluate various energies involved in a process No information about direction of process can be obtained by application of the first law Figure: Heat Flow Direction ELO 2.1

Areas of application of second law is study of energy-conversion systems The second law can be used to derive an expression for the maximum possible energy conversion efficiency taking those losses into account The second law denies the possibility of completely converting into work all of the heat supplied to a system operating in a cycle ELO 2.1

The Second Law is used to determine the maximum efficiency of any process A comparison can then be made between the maximum possible efficiency and the actual efficiency obtained Figure: Rankine Cycle for a Typical Steam Plant ELO 2.1

Entropy Entropy helps explain the Second Law of Thermodynamics
A measure of the unavailability of heat to perform work in a cycle Change in entropy determines the direction in which a given process proceeds The second law predicts that not all heat provided to a cycle can be transformed into an equal amount of work. Some heat rejection must take place In a heat transfer process, minimizing the temperature at which the process occurs will reduce the change in entropy. For example, the greater the heat transfer surface area, the smaller the Delta-T. The better the overall heat transfer coefficient (U), the lower the Delta-T. (Q-dot = UA Delta-T) Figure: Entropy ELO 2.1

Entropy Change in entropy is the ratio of heat transferred during a reversible process to the absolute temperature of the system ∆𝑆= ∆𝑄 𝑇 𝑎𝑏𝑠 (𝑓𝑜𝑟 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠) ∆𝑆 = the change in entropy of a system during some process (𝐵𝑇𝑈/°𝑅) ∆𝑄 = the amount of heat added to the system during the process (BTU) 𝑇𝑎𝑏𝑠 = the absolute temperature at which the heat was transferred (°𝑅) The second law can also be expressed as ΔS ≥ 0 for a closed cycle. Entropy must increase or stay the same It can never decrease ELO 2.1

Entropy Entropy is a extensive property of a system and may be calculated from specific entropies (S = ms) Specific entropy (s) values are tabulated specific enthalpy and specific volume in the steam tables Specific entropy is a property and is used as one of the coordinates when representing a reversible process graphically Area under a reversible process curve on T-s diagram represents the quantity of heat transferred during the process Figure: T-s Diagram With Rankine Cycle ELO 2.1

Entropy Note the “real” process on the T-s diagram results in more area under the curve Thermodynamic problems, processes, and cycles are often investigated by substituting reversible processes for the actual irreversible process Helpful because only reversible processes can be depicted on the diagrams Actual or irreversible processes cannot be drawn since they are not a succession of equilibrium conditions Only the initial and final conditions of irreversible processes are known Figure: T-s and h-s Diagrams for Expansion and Compression Processes ELO 2.1

Entropy Knowledge Check
The second law of thermodynamics can also be expressed as ________ for a closed cycle. Sf = Si ΔS ≥ 0 ΔT > 0 ΔS < 0 Correct answer is B. Correct answer is B ELO 2.1

Carnot’s Principle ELO Given a thermodynamic system, determine the: Maximum efficiency of the system, Efficiency of the components within the system. Related KAs K1.02 Define thermodynamic cycle efficiency in terms of net work produced and energy applied. 1.6* 1.8*; K1.08 Explain the difference between real and ideal turbine efficiency. 1.6* 1.7* Figure: Carnot Cycle Representation ELO 2.2

Carnot’s Principle No engine can be more efficient than a reversible engine operating between the same high temperature and low temperature reservoirs Efficiencies of all reversible engines operating between the same constant temperature reservoirs are the same Efficiency of a reversible engine depends only on the temperatures of the heat source and heat receiver Figure: Carnot’s Efficiency Principle ELO 2.2

Carnot Cycle Reversible isothermal expansion (1-2: TH = constant)
Reversible adiabatic expansion (2-3: Q = 0, TH→TL) Reversible isothermal compression (3-4: TL = constant) Reversible adiabatic compression (4-1: Q = 0, TL→TH) Note that the “net effect” in the change in specific entropy shown in the T-s diagram for this cycle equals ZERO. Figure: Single Piston Carnot Engine Cycle ELO 2.2

Carnot Cycle Heat added (QH) is the area under line 2-3
Heat rejected (QC) is the area under line 1-4 Difference between heat added and heat rejected is the net work (sum of all work processes), which is represented as the area of rectangle The cycle consists of the following reversible processes. 1-2: adiabatic compression from Tcold to Thot due to work performed on fluid. 2-3: isothermal expansion as fluid expands when heat is added to the fluid at temperature TH. 3-4: adiabatic expansion as the fluid performs work during the expansion process and temperature drops from TH to TC. 4-1: isothermal compression as the fluid contracts when heat is removed from the fluid at temperature TC. Figure: Carnot Cycle Representation ELO 2.2

Carnot Cycle Shows max possible efficiency exists when either
TH is at its highest possible value TC is at its lowest possible value Carnot efficiency Represent reversible processes Upper limit of efficiency for any given system operating between the same two temperatures However, all practical/real systems and processes are irreversible and the actual system will not reach this efficiency value ELO 2.2

Carnot Cycle Efficiency (η) of the cycle is the ratio of the net work of the cycle to the heat input to the cycle. Expressed by equation: 𝜂= 𝑄 𝐻 − 𝑄 𝐶 𝑄 𝐻 = 𝑇 𝐻 − 𝑇 𝐶 𝑇 𝐻 =1− 𝑇 𝐶 𝑇 𝐻 Where: 𝜂 = cycle efficiency 𝑇 𝐶 = designates the low-temperature reservoir (˚R) 𝑇 𝐻 = designates the high-temperature reservoir (˚R) °R must be used in calculation ELO 2.2

Carnot Cycle The most important aspect of the second law is the determination of maximum possible efficiencies obtained from a power system Actual efficiencies will always be less than this maximum System losses (friction, windage, turbulence, etc.) Systems are not reversible ELO 2.2

Carnot Cycle Example: Actual Versus Ideal Efficiency
The actual efficiency of a steam cycle is 18.0%. The facility operates from a steam source at 340 °F and rejects heat to atmosphere at 60 °F. Compare the Carnot efficiency to the actual efficiency. 35% as compared to 18.0% actual efficiency ELO 2.2

Carnot Cycle Example: Actual Versus Ideal Efficiency
Figure: Real Process Cycle Compared to Carnot Cycle ELO 2.2

Carnot Cycle An open system analysis was performed using the First Law of Thermodynamics Second law problems are treated the same way An isolated, closed, or open system used in the analysis depends on the types of energy that cross the boundary Open system analysis using the second law equations is the more general case Closed and isolated systems are special cases of the open system ELO 2.2

Carnot Cycle Fluid moves through control volume while work is delivered external to the control volume Assume the control volume boundary is at some environmental temperature and that all of the heat transfer (Q) occurs at this boundary Entropy is a property May be transported with the flow of the fluid into and out of the control volume Figure: Control Volume for Second Law Analysis ELO 2.2

Carnot Cycle Entropy flow into the control volume resulting from mass transport is 𝑚 𝑖𝑛 s 𝑖𝑛 Entropy flow out of control volume is 𝑚 𝑜𝑢𝑡 s 𝑜𝑢𝑡 Assuming that properties are uniform at sections in and out Entropy may also be added to control volume because of heat transfer at control volume boundary Figure: Control Volume for Second Law Analysis ELO 2.2

Carnot Cycle The Second Law of Thermodynamics gives an upper limit for how efficiently a thermodynamic system can perform Determining that efficiency: Know the inlet and exit temperatures of the overall system Apply Carnot’s efficiency equation using those temperatures in absolute degrees ELO 2.2

Carnot Cycle Knowledge Check
The theoretical maximum efficiency of a steam cycle is given by the equation: Effmax = (1 - Tout/Tin) x 100% where Tout is the absolute temperature for heat rejection and Tin is the absolute temperature for heat addition. (Fahrenheit temperature is converted to absolute temperature by adding 460°F.) A nuclear power plant is operating with a stable steam generator pressure of 900 psia. What is the approximate theoretical maximum steam cycle efficiency this plant can achieve by establishing its main condenser vacuum at 1.0 psia? 35 percent 43 percent 65 percent 81 percent Correct answer is B. Correct answer is B NRC Bank Question – P2778 Analysis: To calculate maximum theoretical thermal efficiency, one must determine the two temperatures, convert to degrees Rankine, and solve the given equation: At 900 psia, saturation temperature is 532ºF. Adding 460º, this equals 992 R. This is Tin; the steam generators are the input to the system. At 1 psia, saturation temperature is 102ºF. Adding 460º, this equals 562 R. This is Tout. Effthmax = (1 - Tout/Tin) X 100% = (1 – 562 / 992) x 100% = 43% ELO 2.2

Thermodynamics of Ideal and Real Processes
ELO Differentiate between the paths for an ideal process and for a real process on a T-s or h-s diagram Related KA K1.01 Explain the relationship between real and ideal processes. 1.8* 1.9*; K1.02 Explain the shape of the T-s diagram process line for a typical secondary system. 1.7* 1.9; Figure: T-s Diagram Shows Paths of Processes ELO 2.3

Diagrams of Ideal and Real Processes
Any ideal thermodynamic process can be drawn on a T-s or an h-s diagram isentropic A real process that approximates the ideal process can be shown Usually with dashed lines Figure: T-s Diagram Shows Paths of Processes ELO 2.3

Diagrams of Ideal and Real Processes
In an ideal process, the entropy will be constant Turbine process is an “expansion” process Real expansion or compression process slants slightly toward the right Entropy increases Smaller change in enthalpy Real processes are irreversible These isentropic processes will be represented by vertical lines on either T-s or h-s diagrams, since entropy is on the horizontal axis and its value does not change Real expansion or compression process slants slightly toward the right. Entropy increases from the start to the end of the process Real expansion process (turbine) results in a smaller change in enthalpy – less work is done by the real turbine Real compression process (pump) - more enthalpy must be added during the pump process - more work done on the system Reversible process indicate a maximum work output for a given input, which can be compared to real processes for efficiency purposes Figure: h-s Diagram Shows Expansion and Compression Paths ELO 2.3

Thermodynamic Power Plant Efficiency
ELO Describe how individual factors affect system or component efficiencies Related KAs K1.01 Explain the relationship between real and ideal processes. 1.8* 1.9*; K1.07 Define turbine efficiency. 1.6* 1.6*; K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency ; K1.04 Describe the moisture effects on turbine integrity and efficiency ; K1.05 State the advantages of moisture separators/repeaters and feedwater heaters for a typical steam cycle ; K1.11 Describe the process of condensate depression and its effect on plant operation 2.4,2.5; K1.12 Explain vacuum formation in condenser processes K1.13 Explain, the condensing process K1.09 Define pump efficiency. 1.3* 1.3*; K1.10 Explain the difference between ideal and real pumping processes * All of the NRC exam bank questions on this chapter are tied to this KA ( K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency ) Figure: Typical Steam Cycle ELO 2.4

Power Plant Components
Three basic types of components in power cycles Turbines Pumps Heat exchangers Each change in the properties of the working fluid It is possible to calculate efficiencies of each individual component Compare actual work to ideal work Figure: Typical Steam Cycle ELO 2.4

Steam Turbine Designed to extract energy from steam and use it to do work in the form of rotating the turbine shaft Steam works as it expands through the turbine The shaft work is then converted to electrical energy by the generator Figure: Turbine Work ELO 2.4

First law, general energy equation - the decrease in the enthalpy of the working fluid Hin - Hout equals the work done by the turbine ( 𝑊 𝑡 ) 𝐻 𝑖𝑛 − 𝐻 𝑜𝑢𝑡 = 𝑊 𝑡 𝑚 ℎ 𝑖𝑛 − ℎ 𝑜𝑢𝑡 = 𝑤 𝑡 Where: Hin = enthalpy of the working fluid entering the turbine (BTU) Hout = enthalpy of the working fluid leaving the turbine (BTU) Wt = work done by the turbine (ft-lbf) 𝑚 = mass flow rate of the working fluid (lbm/hr) hin = specific enthalpy of the working fluid entering the turbine (BTU/lbm) hout = specific enthalpy of the working fluid leaving the turbine (BTU/lbm) 𝑤 𝑡 = power of turbine (BTU/hr) ELO 2.4

An ideal turbine provides a basis for analyzing the performance of turbines An ideal turbine performs the maximum amount of work theoretically possible for the plant’s operating conditions 𝑆 𝑖𝑛 = 𝑆 𝑜𝑢𝑡 𝑠 𝑖𝑛 = 𝑠 𝑜𝑢𝑡 Where: 𝑆 𝑖𝑛 = entropy of the working fluid entering the turbine (BTU/°R) 𝑆 𝑜𝑢𝑡 = entropy of the working fluid leaving the turbine (BTU/°R) 𝑠 𝑖𝑛 = specific entropy of the working fluid entering the turbine (BTU/lbm-°R) 𝑠 𝑜𝑢𝑡 = specific entropy of the working fluid leaving the turbine (BTU/lbm-°R) Figure: Mollier Diagram ELO 2.4

Ideal versus Real Turbine KE and PE changes and heat lost by the working fluid in the turbine are negligible In an ideal case, the working fluid does work reversibly by expanding at a constant entropy Entropy of the working fluid entering the turbine: Sin equals the entropy of the working fluid leaving the turbine Sout Figure: h-s Diagram for Ideal and Real Turbine ELO 2.4

An actual turbine does less work because of Friction losses in the blades Leakage past the blades Mechanical friction Turbine efficiency (ηt) is the ratio of the actual work done by the turbine Wt.actual to the work that would be done by the turbine if it were an ideal turbine Wt.ideal ELO 2.4

Turbine Efficiency 𝜂 𝑡 = 𝑊 𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊 𝑡.𝑖𝑑𝑒𝑎𝑙
𝜂 𝑡 = 𝑊 𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊 𝑡.𝑖𝑑𝑒𝑎𝑙 𝜂 𝑡 = ℎ 𝑖𝑛 − ℎ 𝑜𝑢𝑡 𝑎𝑐𝑡𝑢𝑎𝑙 ℎ 𝑖𝑛 − ℎ 𝑜𝑢𝑡 𝑖𝑑𝑒𝑎𝑙 Where: 𝜂 𝑡 = turbine efficiency (no units) Wt.actual = actual work done by the turbine (ft-lbf) Wt.ideal = work done by an ideal turbine (ft-lbf) (hin – hout)actual = actual enthalpy change of the working fluid (BTU/lbm) (hin – hout)ideal = enthalpy change of the working fluid in an ideal turbine (BTU/lbm) Even though this equation is NOT on the NRC Equation Sheet, the only bank question that requires it use has a derivation of this formula provided in the STEM of the question. ELO 2.4

Turbine Efficiency Turbine efficiency is generally
60% to 80% for small turbines 90% for large turbines Turbine efficiency ηt is normally given by the manufacturer Permits actual work done to be calculated directly by multiplying turbine efficiency ηt by work done by an ideal turbine under the same conditions Figure: Entropy Diagram Measures System Efficiency ELO 2.4

Turbine Efficiency Ideal case is a constant entropy represented by a vertical line Actual turbine involves an increase in entropy The smaller the increase in entropy, the closer the turbine efficiency ηt is to 1.0 or 100% Figure: Entropy Diagram Measures System Efficiency ELO 2.4

Real vs. Ideal Cycle Efficiency
Efficiency of a Carnot cycle solely depends on the temperature of the heat source and the heat sink To improve efficiency increase the temperature of the heat source and decrease the temperature of the heat sink For a real cycle the heat sink is limited by the fact that the earth is our final heat sink, and therefore, is fixed at about 60 °F (520 °R) Most plants notice an increase in plant efficiency when the heat sink temperature drops in the winter ELO 2.4

Heat source is limited to the combustion temperatures of the fuel to be burned Fossil fuel cycles upper limit is ~ 3,040 °F (3,500 °𝑅) Not attainable due to the metallurgical restraints of the boilers, Limited to about 1,500 °F (1,960 °𝑅) for a maximum heat source temperature Using these limits to calculate the maximum efficiency attainable by an ideal Carnot cycle gives the following: 𝜂= 𝑇 𝑆𝑂𝑈𝑅𝐶𝐸 − 𝑇 𝑆𝐼𝑁𝐾 𝑇 𝑆𝑂𝑈𝑅𝐶𝐸 = 1,960 °𝑅 −520 °𝑅 1,960 °𝑅 =73.5% Our actual steam temperature (heat source) is based on saturation pressure that the SG is at. For a pressure of 1000 psia, Tsat is about 545oF, which equates to 1005oR. ELO 2.4

Heat Rejection 73% efficiency is not possible Energy added to a working fluid during the Carnot isothermal expansion is given by qs Not all of this energy is available for use by the heat engine since a portion of it (qr) must be rejected to the environment ELO 2.4

𝑞 𝑟 = 𝑇 𝑜 𝛥𝑠 Measured in units of BTU/lbm Where: To = average heat sink temperature of 520 °R Available Energy (A.E.) for Carnot cycle Equation: 𝐴.𝐸. = 𝑞𝑠 − 𝑞𝑟 Substituting equation for qr gives: 𝐴.𝐸. = 𝑞𝑠 − 𝑇0 Δ𝑠 Between the temperatures 1,960 °R and 520 °R ELO 2.4

Operating less than 1,962 °R is less efficient The change in entropy corresponds to a measure of the heat rejected by the engine Recall: Entropy is a measure of the energy unavailable to do work Developing materials capable of withstanding the stresses above 1,962 °R increases the energy available for use by the plant cycle Figure: Entropy Measures Temperature as Pressure and Volume Increase ELO 2.4

Actual available energy (area ) Less than half of the ideal Carnot cycle (area 1-2'-4) operating between the same two temperatures Fossil plants are about 40% efficient Nuclear plants are around 31% Fossil fuel power cycle – 2,000 psia maximum, heat addition at temperature of 1,962 °𝑅 is not possible Water requires heat addition in a constant pressure process Average temperature for heat addition is below the maximum 1,962 °𝑅 Figure: Carnot Cycle Versus Typical Power Cycle Available Energy ELO 2.4

Carnot steam cycle on a T-s diagram A great deal of pump work is required to compress a two phase mixture of water and steam from point 1 to the saturated liquid state at point 2 Pump cavitation would exist Condenser designed to produce a two-phase mixture at the outlet Would pose technical problems Figure: Ideal Carnot Cycle ELO 2.4

Ideal Rankine Cycle QA – red + green area QR – green area
WT – red area The greater the QA and the less the QR The greater the WT Operational analogy: The greater the steam pressure, and, The lower the backpressure The more work of the turbine Figure: Rankine Cycle An example shown on the next slide. ELO 2.4

Figure: Rankine Cycle Efficiency Comparisons on a T-s Diagram
Ideal Rankine Cycle Rankine Cycle Efficiencies Cycle A has less heat rejected than Cycle B Better vacuum, for example Both have same amount of heat added Cycle A Same heat added Less heat rejected More work of turbine More efficient Figure: Rankine Cycle Efficiency Comparisons on a T-s Diagram ELO 2.4

Real vs Ideal Rankine Cycle
The ideal turbine replaced with a real turbine The efficiency is reduced The non-ideal turbine incurs an increase in entropy The increase in the area of 3-2-3‘ < the increase in area of a-3-3'-b Also, condensate slightly subcooled More sensible heat added to FW Green line For simplification this diagram only showed the difference between IDEAL and REAL for the turbine process. The IDEAL vs REAL Pump process has been shown in an earlier slide. The added green line shows the REAL pump process (slight change in specific entropy). Even though it raises the specific enthalpy of the fluid (higher temperature of fluid), it still requires a larger pump to get the required pressure increase to make up for the headloss. Figure: Rankine Cycle With Real Versus Ideal Turbine ELO 2.4

Rankine Cycle – Steam Plant
Processes that comprise the steam cycle: 1-2: Heat added in the SG under a constant pressure 2-3: Saturated steam is expanded in high pressure (HP) turbine to provide shaft work 3-4: HP exhaust is dried and superheated in the MSR Figure: Typical Steam Cycle ELO 2.4

4-5: Superheated steam in LP turbine provides shaft work 5-6: Exhaust from the turbine is condensed under a constant vacuum 6-7: Condensate is compressed as a liquid by the condensate pump Figure: Typical Steam Cycle ELO 2.4

7-8: Condensate is preheated by the low pressure feedwater heaters 8-9: Condensate is compressed as a liquid by the feedwater pump 9-1: Feedwater is preheated by the high- pressure heaters 1-2: Cycle starts again Figure: Typical Steam Cycle ELO 2.4

Ideal Rankine Cycle – Steam Plant
Differences between IDEAL and REAL: Ideal pumps and turbines do not exhibit an increase in entropy; real pumps and turbines will No condensate subcooling as point 6 is on the saturation line; condensate actually slightly subcooled (condensate depression) Figure: Typical Steam Cycle Figure: Rankine Steam Cycle (Ideal) ELO 2.4

Rankine Cycle Steam Plant – Real T-s
Additional heat rejected in condenser must then be made up for in the SG Real pumps and turbines would exhibit an entropy increase across them Subcooling decreases cycle efficiency, but aids in preventing condensate pump cavitation Keep in mind that from a cycle efficiency standpoint, subcooling is BAD. For a given amount of heat available from the RCS to be added to the secondary, the lower the temperature of the feedwater entering the SG, the lower the temperature (pressure) exiting the SG. A lower steam pressure means less change in specific enthalpy of the turbine. This can be shown on a Mollier Diagram. Figure: Steam Cycle (Real) ELO 2.4

MSR Effect on Turbine Work
Point 1: Saturated steam at 540 °F Point 2: 82.5% quality at exit of HP turbine exhaust Point 3: Temperature of superheated steam is 440 °F Point 4: Condenser vacuum in psia Points 1-2: HP turbine ideal work Points 3-4: LP turbine ideal work NOTE: Efficiency can increase or decrease with MSRs, but the main reason for using MSRs is to minimize impingement in the final stages of the LP turbine blading. Generically, the NRC looks at the use of the MSR as an “increase” in efficiency. Figure: Mollier Diagram ELO 2.4

Causes of Inefficiency
Component Examples Turbine efficiency limited by: Friction between the steam and the turbine surfaces Friction between the steam and entrained liquid Drag from impingement of moisture on rotating turbine parts Friction between the rotating and stationary parts of the turbine Throttling of steam flow to the turbine Steam leakage around fixed and moving blades Pumps Fluid and mechanical friction Condenser Non-condensable gasses (lower vacuum) ELO 2.4

Causes of Inefficiency
Cycles Condensers are designed to subcool the liquid by 8-10°F Allows the condensate pumps to pump without cavitation Requires more sensible heat to be added Ambient losses Again, this is energy lost to the system and therefore unavailable to do work ELO 2.4

Improved Cycle Efficiency
Condition Effect Discussion Superheating More Efficient With More Superheating Increased heat added results in more net work from the system, even though more heat is rejected. Moisture Separator Reheater (MSR) Use of MSR Has Minor Effect On Efficiency More work is done by the low-pressure (LP) turbine since inlet enthalpy is higher but more heat is rejected. The principle benefit of MSR use is protection of the final blading stages in LP turbine from erosion by water droplet impingement. Feedwater Preheating More Efficient With Feedwater Preheating Less heat must be added from the heat source (reactor) since the feedwater enters the steam generator closer to saturation temperature. Condenser Vacuum More Efficient With Higher Vacuum (Lower Backpressure) Net work output is higher and heat rejection is lower as condenser pressure is lowered. Keep in mind that some of these “conditions” come at a cost, but generically they ALL result in an increase in cycle efficiency. For example: The MSR’s raise the enthalpy of steam entering the LP turbine which means more work of the turbine. However, steam going to the MSR from the SG does not go to the HP turbine (less mass flow rate) Feedwater heating raises the temperature of the “pump” process meaning less sensible heat must be added in the SG. However, since this uses extraction steam, there is less steam mass flow rate going through the LP and HP turbines. ELO 2.4

Improved Cycle Efficiency
Condition Effect Discussion Condensate Depression More Efficient With Minimal Condensate Depression Minimal condensate depression reduces both the amount of heat rejected and the amount of heat that must be supplied to the cycle. Steam Temperature/ Pressure More Efficient At Higher Steam Temperature/ Pressure At higher steam temperature, the inlet and exit entropy from the turbine are lower so less heat is rejected. Steam density increases as pressure increases, so more turbine work is done. Steam Quality More Efficient At Higher Steam Quality Enthalpy content increases as moisture content decreases and more net work is done. ELO 2.4

Knowledge Check To achieve maximum overall nuclear power plant thermal efficiency, feedwater should enter the steam generator (SG)__________ and the pressure difference between the SG and the condenser should be as __________ as possible. close to saturation; great close to saturation; small as subcooled as practical; great as subcooled as practical; small Correct answer is A. Correct answer is A. NRC Bank Question – P478 Analysis: Condensate depression decreases the overall plant efficiency because more sensible heat must be added to reach saturated conditions in the steam generators. However, the advantage of having a small degree of condensate depression is to ensure adequate net positive suction head (NPSH) for the main condensate pumps and prevent cavitation. Excessive condensate depression causes more air absorbed in condensate and accelerates oxygen corrosion of secondary plant materials. Therefore, to maximize plant efficiency, feedwater should enter the steam generator as close to saturation as possible. If the water were highly subcooled, more sensible heat must be added to reach saturated conditions in the steam generators. Having a large pressure difference between the steam generator and condenser enhances thermal efficiency. For a given Mollier diagram, note the increased turbine work (due to larger change in enthalpy) for the condenser at a 28” Hg vacuum (1 psia absolute) compared to the condenser at atmospheric pressure (14.7 psia). Also note, that as you continue to increase steam pressure (after the hump around 500 psia), the specific enthalpy decreases slightly. However, based on the slope of the backpressure line the change in specific enthalpy increases as SG pressure increases. ELO 2.4

Knowledge Check A nuclear power plant was initially operating at steady-state 90 percent reactor power when extraction steam to the feedwater heaters was isolated. With extraction steam still isolated, reactor power was returned to 90 percent and the plant was stabilized. Compared to the initial main generator MW output, the current main generator MW output is... lower, because the steam cycle is less efficient. higher, because the steam cycle is less efficient. lower, because more steam heat energy is available to the main turbine. higher, because more steam heat energy is available to the main turbine. Correct answer is A. Correct answer is A. NRC Bank Question – P3378 Analysis: Isolating steam to a high pressure feedwater heater will lower plant efficiency; the feedwater will now enter the steam generator at a lower temperature (further from saturated conditions). More sensible heat must be added to raise the feedwater to saturated conditions. Recall that efficiency of the plant is equal to the work of the turbine divided by the heat added from the RCS (Nplant = Mwe/Mwth). Since losing feedwater heating is a loss of cycle efficiency, if reactor power is returned back to original value, Turbine power must be lower. ELO 2.4

NRC KA to ELO Tie KA # KA Statement RO SRO ELO K1.01
Define thermodynamic cycle. 1.6 1.7 2.1 K1.02 Define thermodynamic cycle efficiency in terms of net work produced and energy applied. 1.8 2.2, 2.3 K1.03 Describe how changes in secondary system parameter affect thermodynamic efficiency. 2.5 2.6 2.4 K1.04 Describe the moisture effects on turbine integrity and efficiency. 2.3 K1.05 State the advantages of moisture separators/repeaters and feedwater heaters for a typical steam cycle. 1.9

Operator Generic Fundamentals – Thermodynamic Cycles

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